Word Problems

Let the initial number of cookies be x
Total weight of the mixture = 600 ounces

Since they are identical the weight per cookie will be the same
Weight per cookie is \(\frac{600}{x}\)

Let the new cookies be y (cookies produced after reducing the weight by 1 ounce)
Given that they produced 30 cookies more than the previous batch
Hence

y=30+x

y-x=30

Total weight of the mixture = 600 ounces

New weight per cookie after reducing the weight = \(\frac{600}{y}\)

Given that the difference in weight is 1 ounce

or

\(\frac{600}{x}\) - \(\frac{600}{y}\) =1

Simplifying we get

\(\frac{600\left(y-x\right)}{xy}\) =1

We know that y-x=30, substituting we get

\(\frac{18000}{xy}\) =1

xy=18000

Returning back to our initial equation that is

y-x=30

Multiplying it by x on both sides we get

\(xy\ -\ x^2=30x\)

Substituting the value of xy and rearranging we get

\(x^2+30x\ -\ 18000\) =0

Solving the quadratic

\(x^2+150x\ -\ 120x\ -\ 18000\) = 0

\(x\left(x+150\right)\ -\ 120\left(x\ +\ 150\right)=0\)

\(\left(x+150\right)\left(x\ -\ 120\right)=0\)

x=-150 or x=120

Since the number of cookies can't be negative we get that the initial mixture had 120 cookies

Weight per cookie in the initial case = \(\frac{600}{120}=5\)